elasticityhomework-and-exercises
I always thought that the elastic limit is at an exact point for a force extension graph. I do not understand why the elastc limit is beyond point S. I also know that the elastic limit means the max force a wire can handle allowing it to return to its original shape.
(This explanation is adapted from Nicholas Wheeler Notes, nevertheless is self-contained, also a slightly modified version is published on my website A Sudden Burst of Physics, Math and more ):
One can clearly see how a quantity like $\mu = m/a$ (mass density per unit length) would yield a finite result in the refinement process since one expects both the mass $m$ and the lattice spacing $a$ to decrease when we go to very small length-scales.
However for the quantity $ka$ to yield a finite result, each spring proper stiffness $k$ would become necessarily stronger and stronger as the lattice refinement process proceeds: $k(a) \nearrow \infty$ as $a \searrow 0$. This is the problematic concept !!, however as we'll see in the refinement process, the springs $k$ don't add up cumulatively to an effective constant $K$, but they add in series (like parallel resistors) as opposed to our intuition.
To see this lets consider the case of a finite spring chain with stiffness $k$ and $N$ masses,
Clearly, if the total length of the spring chain (length between the two barriers) is $l$ and the length between the masses in equilibrium is $a$, then we have $$(N+1)a = l, \qquad M = Nm,$$
where $M$ is the total mass of the spring chain.
We can rewrite the previous expressions as,
$$N = \frac{l}{a}\left(1-\frac{a}{l}\right) , \qquad m = \frac{M}{l}a\left(1-\frac{a}{l}\right)^{-1}= \mu a\left(1-\frac{a}{l}\right)^{-1}, $$
where $\mu=M/l$ is the linear mass density of the spring chain.
So what we are going to impose is that in the refinement $a \searrow 0$ the quantity $\mu$ keeps constant, the same in the limiting case of a compressional wire (or "string") as for the spring chain from which we started. This implies that the total number of springs $N$ have to increase like $\mathcal{O}(a^{-1})$ and the masses $m$ of each of them have to decrease as $\mathcal{O}(a)$. To see this clearly when $a\ll l$ in the previous equations,
$$N = \frac{l}{a}, \qquad m = \mu a. $$
From this equations, we can see that since $\mu$ is kept constant in the limiting process, the value of the individual masses $m$ are going to decrease in the refinement. This allows us to approximate the spring chain when $a\ll l$ as a chain of spring in the so-called series configuration, i.e. a series of springs connected by massless contacts.
So if we have $n$ springs of stiffness $k_i$ each, the series configuration of springs would have a total or effective stiffness $k_T$ of,
$$\frac{1}{k_T}= \frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+...+\frac{1}{k_n}.$$
So if we have a series configuration of springs of length $l$ and effective stiffness $K={Y}/{l}$ that have been assembled by connecting in series $N + 1$ identical springs $k$ of length $a=l/(N+1)$ we have,
$$\frac{1}{K}=\frac{l}{Y}=N\frac{1}{k} \qquad \text{with} \qquad N = \frac{l}{a}, \text{ when } a\ll l.$$
Then,
$$k = N \frac{Y}{l} = \frac{Y}{a} \text{ when } a\ll l.$$
This implies that in the refinement process $a \searrow 0$ the quantity,
$$ka \xrightarrow[a \searrow 0]{} Y,$$
where we are imposing $Y$ as a constant in the refinement process.
So we checked our initial claim that the springs $k$ become necessarily stronger and stronger as the lattice refinement process proceeds: $k(a) \nearrow \infty$ as $a \searrow 0$. However since the springs add in series they manage to generate a constant effective spring stiffness $K={Y}/{l}$ in the refinement process, because in the refinement as the spring constant $k$ grows as $\mathcal{O}(a^{-1})$ the number of springs $N$ grows also as $\mathcal{O}(a^{-1})$, then since $K=N/k$, the effective spring stiffness remains constant.
With this result we can take the limit of the potential energy, this yields,
$$U=\frac{1}{2} \sum_i ka \Big(\frac{\phi_{i+1} -\phi_{i}}{a}\Big)^2 a \xrightarrow[a \searrow 0]{} \frac{1}{2}\int Y \Big(\frac{\partial \phi}{\partial x}\Big)^2dx ,$$
where we used
$$a\xrightarrow[a \searrow 0]{}dx \qquad \frac{\phi_{i+1} -\phi_{i}}{a}\stackrel{a \rightarrow \Delta x}{=}\frac{\phi(x+\Delta x)-\phi(x)}{\Delta x}\xrightarrow[\Delta x \searrow 0]{} \frac{\partial \phi}{\partial x}.$$
and that the sum became an integral in the limiting to the continuum.
Gathering all the previous results we obtain the total potential energy:
$$U=\frac{1}{2}\int Y \Big(\frac{\partial \phi}{\partial x}\Big)^2dx.$$
So we obtained the potential energy density
\begin{equation} \frac{dU}{dx}= \frac{1}{2}Y \Big(\frac{\partial \phi}{\partial x}\Big)^2. \end{equation}
Referring to your graph which is for a ductile material I suggest the following.
A is the limit of proportionality up to which the stress and strain are proportional to one another and when unloaded the material goes back to its original length.
B is the elastic limit.
With stresses below this the material behaves elastically i.e., when unloaded returns back to its original length although at the highest stress the graph is no longer a straight line.
It is also true that there is greater increase in strain for a given increase in stress in stress beyond the yield point.
C is the yield point which when reached will result in a permanent deformation of an arbitrarily defined amount of strain (sometimes $0.2\%$) when the stress is removed.
E is the onset of necking (reduction in cross sectional area of the material) where the stress is actually increasing (graph ending at D) but the evaluated stress used for plotting some graphs assumes the cross section of area of the material does not change (graph ending at F).
A really good material to observe these effects is brass in the form of a wire which is about 2 or 3 metres long.
Initially with small loads the wire extends a little bit and elastically.
Then beyond a certain load when an extra weight is added the loads move down visibly (centimetres) and then stops.
The wire is now past its yield point.
This large visible increase in length just after an extra load is added is probably what your teacher was talking about?
Removing the load the wire does not go back to its original length.
Loading the brass wire further produces large visible extensions and then eventually the wire breaks and using a magnifying glass one can easily see the neck which is the characteristic of a ductile fracture.
Best Answer
There's a subtle difference between elastic limit and what is called the limit of proportionality the limit of proportionality is when the graph becomes non linear ( deviates from a straight line / curves) but elastic limit is the point after which when the load is removed the spring does not come back to its original length. So here that point should lie just after S ( as at T it's already far from elastic limit as when the load is removed it's far from original length but just after S there will be a point where after removing load the spring will be just a bit longer than original length) Answer should be B